# How do you solve -\frac { 5} { 3} = \frac { x ^ { 2} - 3} { x + 1}?

Aug 31, 2017

$x = \frac{- 5 + \sqrt{73}}{6} \textcolor{w h i t e}{x} \mathmr{and} \textcolor{w h i t e}{x} x = \frac{- 5 - \sqrt{73}}{6}$

#### Explanation:

$- \frac{5}{3} = \frac{{x}^{2} - 3}{x + 1}$

Following these simple steps accordingly..

Cross Multiply

$- 5 \left(x + 1\right) = 3 \left({x}^{2} - 3\right)$

While the minus $\left(-\right)$ sign is attached to the $5$ is because, in every fraction, the symbol always belongs to the numerator, if the denominator is removes..

Simplify...

$- 5 x - 5 = 3 {x}^{2} - 9$

Collect like terms

$3 {x}^{2} + 5 x - 9 + 5 = 0$

$3 {x}^{2} + 5 x - 4 = 0$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where $a = 3 , \textcolor{w h i t e}{x} b = 5 , \textcolor{w h i t e}{x} c = - 4$

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(3\right) \left(- 4\right)}}{2 \left(3\right)}$

$x = \frac{- 5 \pm \sqrt{25 + 48}}{6}$

$x = \frac{- 5 \pm \sqrt{73}}{6}$

$x = \frac{- 5 + \sqrt{73}}{6} \textcolor{w h i t e}{x} \mathmr{and} \textcolor{w h i t e}{x} x = \frac{- 5 - \sqrt{73}}{6}$